On Sun, 3 Feb 2008 10:20:23 -0800 (PST), kevin cline
<kevin.cline@[EMAIL PROTECTED]
> wrote in
<news:74c154bf-1ed2-408a-b012-4da4ab6592b3@[EMAIL PROTECTED]
>
in alt.algebra.help:
> On Feb 3, 1:06 am, "Brian M. Scott" <b.sc...@[EMAIL PROTECTED]
> wrote:
>> On Sat, 2 Feb 2008 22:50:31 -0800 (PST), kevin cline
>> <kevin.cl...@[EMAIL PROTECTED]
> wrote in
>>
<news:9af82bfd-5b0b-416d-a27c-d3e7d1f1a959@[EMAIL PROTECTED]
>
>> in alt.algebra.help:
>>> On Jan 27, 1:46 pm, "Brian M. Scott" <b.sc...@[EMAIL PROTECTED]
> wrote:
>>>> On Sun, 27 Jan 2008 11:28:59 -0800 (PST), Tallison
>>>> <Tallison.Rau...@[EMAIL PROTECTED]
> wrote in
>>>>
<news:037f548d-334f-43a5-9593-937005da0ff8@[EMAIL PROTECTED]
>
>>>> in alt.algebra.help:
>>>>> The questionut: How many five-card combinations of a standard
playing
>>>>> card deck have cards from exactly two suits?
>>>>> The answer: 379,236.
>>>>> The sad thing is that after working on this problem for the last
hour,
>>>>> I've gotten answers ranging from 32,712 to 65,780 to 268,382.4 (no
>>>>> kidding!) to 328,900 - everything, that is, but the actual answer
>>>>> according to the textbook.
>>>>> So for the love of God, can someone help me identify the correct
>>>>> formula?????
>>>> There are C(4, 2) = 6 ways to choose the two suits to be
>>>> represented in the hand. Now let A and B stand for any two
>>>> of the suits. A 'legal' hand can have:
>>>> (a) 1 card from A and 4 from B;
>>>> (b) 2 cards from A and 3 from B;
>>>> (c) 3 cards from A and 2 from B; or
>>>> (d) 4 cards from A and 1 from B.
>>> Too complicated. Just compute the number of ways to
>>> choose five cards from those two suits and then subtract
>>> the hands where all the cards are in the same suit:
>>> C(26,5) - 2*C(13,5) = 65780 - 2574 = 63206
>> In this problem the two approaches are of roughly comparable
>> difficulty, so the one to be preferred when solving the
>> problem is whichever one you see first. In this case I'd
>> say that it's pretty much a toss-up for paedagogical
>> purposes, too, though I lean towards the more
>> straightforward analysis -- mine -- for beginners.
> Great. Now use your method to compute the number of
> two-suited bridge hands of 13 cards.
How is that relevant to *this* problem, which is explicitly
what I was discussing, both originally and in my response to
you?
Of course there are going to be other problems for which the
straightforward approach is infeasible, or at least very
undesirable. Ideally a student should eventually reach the
point of seeing multiple approaches, but it's necessary to
start somewhere. And it's worth pointing out that some
problems don't have computationally nice solutions. For
example, there are obviously problems of this same general
type for which the direct and complementary approaches are
equally onerous.
(And in fact I did just use the direct approach to calculate
the number of two-suited bridge hands; using a calculator
without built-in binomial coefficients it was only slightly
more annoying than the complementary approach.)
Brian
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